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I believe I need to use the root test and ratio test. I've solved the first two, but I'm not too sure I fully understand how to do these, so was hoping someone else could work them out so I could digest the work.

  1. $\sum_{k=0}^{\infty}\frac{kx^k}{(2k+1)^2}$
  2. $\sum_{k=0}^{\infty}(2+(-1)^k)^kx^{2k}$
  3. $\sum_{k=0}^{\infty}3^{k^2}x^{k^2}$

Work for 1:

$a_k = \frac{k}{(2k+1)^2}$, so $\frac{k}{(2k+1)^2} * \frac{(2(k+1)+1)^2}{k+1}$ by the ratio test approaches 1 as $k \to \infty$, which means $R = 1$. When $x = 1$, the series converges. When $x = -1$, the series is an alternating series, but ultimately approaches 0, so it too is convergent. Hence the interval of convergence is $[-1, 1]$.

Work for 2: $a_k = (2 + (-1)^k)^k$, so by the root test, we get $R = \frac{1}{\lim \sup a_k^{1/k}}$, which ultimately cancels out the outer power of $k$, and hence we need the $\sup$ of $2 + (-1)^k$, which the $\sup$ of would be 3. Hence, $R = \frac{1}{3}$. Then, I think it converges at both endpoints, but I'm not sure and you have an interval of convergence of $[-\frac{1}{3}, \frac{1}{3}]$

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  • $\begingroup$ Perhaps you could edit in your solutions to 1 and 2. $\endgroup$ – Henry Feb 5 '15 at 14:53
  • $\begingroup$ Why don't you post what you have done for the first two so we can go over it. $\endgroup$ – Mattos Feb 5 '15 at 15:00
  • $\begingroup$ In the (2) you forgot a square root. Look at the exponent of the $x$, it is $2k$ instead of $k$. $\endgroup$ – Pp.. Feb 5 '15 at 15:12
  • $\begingroup$ Don't you just ignore $x^{2k}$ because it's not part of $a_k$? $\endgroup$ – David Feb 5 '15 at 15:13
  • $\begingroup$ @David No, we must do exactly what the formula is saying: The $k$-th term with exponent $1/k$. But $(2+(-1)^k)^k$ is not the $k$-th term, it is the $2k$-th term. $\endgroup$ – Pp.. Feb 5 '15 at 15:19
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If $\sum a_kx^k$ is your series.

  1. Ratio test form: $R^{-1}=\limsup_k\frac{|a_{k+1}|}{|a_k|}=\limsup_k\frac{\frac{k+1}{(2(k+1)+1)^2}}{\frac{k}{(2k+1)^2}}=\limsup_k\frac{k+1}{k}\frac{(2k+1)^2}{(2k+3)^2}=1$

  2. Root test form: $R^{-1}=\limsup_k |a_k|^{1/k}=\limsup_k|(2+(-1)^k)^k|^{\frac{1}{2k}}=\sqrt{3}$. Notice the $a_k$'s are sometimes zero, but since it is $\limsup$ we can ignore those terms.

  3. Root test form: $R^{-1}=\limsup_k|a_k|^{1/k}=\limsup_k3=3$. Notice the $a_k$'s are sometimes zero, but since it is $\limsup$ we can ignore those terms.

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  • $\begingroup$ How did you get $\sqrt{3}$ for the second one? Outer $k$ should cancel out and you're left with $2 + (-1)^k$ which I think the sup of is $3$. $\endgroup$ – David Feb 5 '15 at 15:12
  • $\begingroup$ In the (2) you forgot a square root. Look at the exponent of the $x$, it is $2k$ instead of $k$. $\endgroup$ – Pp.. Feb 5 '15 at 15:13
  • $\begingroup$ I thought you ignore it because its not part of $a^k$ $\endgroup$ – David Feb 5 '15 at 15:13
  • $\begingroup$ If your series in (2) is $\sum_k a_kx^k$ then $a_{2k+1}=0$ and $a_{2k}=(2+(-1)^k)^k$ and in the $\limsup$ you get $|a_{2k}|^{1/(2k)}$. $\endgroup$ – Pp.. Feb 5 '15 at 15:16
  • $\begingroup$ In 3 do you do the root test twice? Going from $3^{k^2}$ to $3^k$, and then ultimately $3$? $\endgroup$ – David Feb 5 '15 at 15:24

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